Problem: Let $f(x)=\sqrt{x^3}$. $f'(16)=$
Answer: The strategy We can first rewrite $f(x)$ as a rational power of $x$. Then, the derivative of $f$ can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is a fraction.) Once we have $f'(x)$, we can plug $x=16$ into it to find $f'(16)$. Rewriting the radical as a rational power $f(x)=\sqrt{x^3}=x^{^{\frac{3}{2}}}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}\left(x^{^{\frac{3}{2}}}\right) \\\\ &=\dfrac{3}{2}x^{^{\frac{3}{2}-1}} \gray{\text{The power rule}} \\\\ &=\dfrac32x^{^{\frac{1}{2}}} \end{aligned}$ Evaluating $f'(x)$ So we found that $f'(x)=\dfrac32x^{^{\frac{1}{2}}}$, which can also be written as $1.5\sqrt{ x}$. Now let's plug ${x=16}$ : $\begin{aligned} 1.5\sqrt{{16}}&=1.5\cdot 4 \\\\ &=6 \end{aligned}$ In conclusion, $f'(16)=6$.